Q13 / 200 · Manipal (MET) 2014 · Chemistry
KBr crystallises in NaCl type of crystal lattice and its density is $2.75 \mathrm{~g} / \mathrm{cm}^3$. Number of unit cells of KBr present in a $1.00 \mathrm{~mm}^3$ grain of KBr are
- Correct answerA$3.6 \times 10^{18}$
- B$3.6 \times 10^{16}$
- C$2.34 \times 10^{23}$
- D$6.02 \times 10^{24}$
Solution
In a unit cell of KBr , there are four formula units of KBr therefore,
$\text { Density }(\rho)=\frac{4 \times 119}{6.023 \times 10^{23} \times a^3}=2.75$
$\begin{aligned}
a^3 & =2.873 \times 10^{-22} \mathrm{~cm}^3 \\
& =2.778 \times 10^{-19} \mathrm{~mm}^3
\end{aligned}$
Number of unit cells per $\mathrm{mm}^3$ $=\frac{10^{19}}{2.873}=3.5 \times 10^{18}$
Number of unit cells per $\mathrm{mm}^3$ $=\frac{10^{19}}{2.873}=3.5 \times 10^{18}$
Hence, the correct answer is option: 'A'.